ee 230 lecture 12 spring 2010 - iowa state universityclass.ece.iastate.edu/ee230/lectures/ee 230...

EE 230Lecture 12

Basic Feedback ConfigurationsGeneralized Feedback Schemes

IntegratorsDifferentiatorsFirst-order active filtersSecond-order active filters

Input and Output Impedances with Feedback

RINF=?

ROF=?

Exact analysis :

1

1 2

Rβ =R +R

AVV1V1

Two-Port Nonideal Op Amp

RIN

RO

R1

R2

Consider amplifier as a two-port and use open/short analysis method

Will find RINF, ROF, AV almost identical to previous calculationsWill see a small AVR present but it plays almost no role since RINF is so large (effectively unilateral)

VF1Aβ

00F

V

RR1+βA ( )1INF IN VR =R A β+ VRFA β

Review from Last Time

Buffer Amplifier

1OUTV

IN

VAV

= =

Provides a signal to a load that is not affected by a source impedance

One of the most widely used Op Amp circuits

Special case of basic noninverting amplifier with R1=∞ and R2=0 1OUT 2

IN 1

V RV R

= +

RIN=∞

ROUT= 0

This provides for decoupling between stages in many circuits


Basic Inverting Amplifier

OUT 2

IN 1

V R-V R

=

2

1

IN

OUT

RIN=R1

OUTIN

1 2

VV + =0R R

ROUT= 0

Input impedance of R1 is unacceptable in many (but not all) applications

This is not a voltage feedback amplifier (it is a feedback amplifier) of the type (note RIN is not high!)

Feedback concepts could be used to analyze this circuit but lots of detail required


Summing AmplifierF

11

OUT2

2

kk

OUT 1 2 k

F 1 2 k

V V V V+ + +...+ =0R R R R

F F FOUT 1 2 k

1 2 k

R R RV = - V - V -...- VR R R

• Output is a weighted sum of the input voltages• Any number of inputs can be used• Gains from all inputs can be adjusted together with RF• Gain for input Vi can be adjusted independently with Ri for 1 ≤ I ≤ k• All weights are negative• Input impedance on each input is Ri


Generalized Inverting Amplifier

( ) ( )( )

F

1

Z sT s

Z s= −

s-domain representation

Z1 and ZF can be any s-domain circuits

If Z1=R, ZF=1/sC, obtain ( ) 1T ssRC

= −

What is this circuit?


( ) 1T ssRC

= −


Consider the differential equation ( )0

t

y K x dτ τ= ∫

( ) ( )X sY s = K

sTaking the Laplace Transform, obtain ( ) ( )

( )Y s KT s = = X s s

Thus, this circuit is an inverting integrator with a unity gain frequency of K = (RC)-1

K is the frequency where |T(jω)|=1 and is termed the Integrator Unity Gain Frequency

Inverting Integrator

( ) 1T ssRC

= −

( ) 1T jωjωRC

= −

( ) 1T jωωRC

=

( ) 90T jω o∠ =

Unity gain frequency is 01ω

RC=


( ) 1T ssRC

= −

( ) ( )0

OUT IN IN1V V V 0

RC

t

dτ τ= − +∫Integrators are widely used !

The integrator function itself is ill-conditioned and integrators are seldom used open-loop

If the input has any dc component present, since superposition applies, the output would diverge to ±∞ as time increases

The offset voltage (discussed later) will also cause an integrator output to diverge

The ideal integrator has a pole at s=0 which is not in the LHP

RIN=R ROUT=0

Inverting Integrator( ) ( )

0OUT IN IN

1V V V 0RC

t

dτ τ= − +∫

What is the output of an ideal integrator if the input is an ideal square wave?

Amplitude of output dependent upon RC product

Inverting Integrator( ) ( )

0OUT IN IN

1V V V 0RC

t

dτ τ= − +∫

What is the output of an ideal integrator if the input is an ideal sine wave?

Amplitude of output dependent upon RC product


( ) ( )0

OUT IN IN1V V V 0

RC

t

dτ τ= − +∫

If VIN(0) = 0,

Ill-conditioned nature of open-loop integrator

0OUT DC

1V VRC

t

dτ= − ∫

0

DCOUT

VV 1RC

t

dτ= − ∫

( )0tDC DCOUT

V VV tRC RC

τ= − = −

For any values of VDC, R, and C, the output will diverge to ± ∞

DC

OUT


( ) ( )0

OUT IN IN1V V V 0

RC

t

dτ τ= − +∫

Ill-conditioned nature of open-loop integrator

( ) ( )1

IN 0 kV t A sin kωt+θk

∞

=

= +∑

Any periodic input that in which the average value is not EXACTLY 0 will have a dc component

This dc input will cause the output to diverge!

Noninverting Integrator

( ) ( )0

OUT IN IN1V V V 0

RC

t

dτ τ= +∫

Obtained from inverting integrator by preceding or following with inverter

Also widely used

Same issues affect noninverting integrator

Requires more components

Lossy Integrator

IN

OUT

F

Add a large resistor to slowly drain charge off of C and prevent divergence

Allows integrator to be used “open-loop”

Changes the dc gain from -∞ to -RF/R

But the lossy integrator is no longer a perfect integrator

What if RF is not so large?

IN

OUT

F

( ) ( )( )

F

1

Z sT s

Z s= −

What if RF is not so large?

( ) ( )( )

F

1

Z sT s

Z s= −

( )

2

2

1

1RsC

1R +sCT s

R

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠= − 2

1 2

R 1R 1+sCR

⎛ ⎞= −⎜ ⎟

⎝ ⎠

First-order lowpass filter with a dc gain of R2/R1

( ) 2

1 2

R 1T sR 1+sCR

⎛ ⎞= −⎜ ⎟

⎝ ⎠( )T jω

2

1R C

2

1

RR

R2 controls the pole (and also the dc gain)

R1 controls the dc gain(and not the pole)

Summing Integrator

R1V1 VOUT

R2V2

RkVk

C

( ) 1T ssRC

= −

OUT IN1V V

sRC= −

By superposition .....OUT 1 2 k1 2 k

1 1 1V V V VsR C sR C sR C

= − − − −

• All inverting functions• Can have any number of inputs• Weights independently controlled by resistor values• Weights all changed by C

1OUT i

i

1V VsR C

k

i=

= −∑

R2

R1

VOUT

VIN

R

VIN

VOUT

C

OUT

IN

OUT 2

IN 1

V R= -V R

OUT 2

IN 1

V R= 1+V R

OUT

IN

V 1= -V sRC

OUT

IN

V 1= +V sRC

?

I’ve got a better noninverting integrator !

OUT

INOUT

IN

V 1= +V sRC

?

I’ve got a better noninverting integrator !

( )IN OUTV sC+G =V sC

OUT

IN

V 1+sRC=V sRC

It has a noninverting transfer function

But it is not an noninverting integrator !

But is this a useful circuit?

OUT

IN

This is a first-order high-pass amplifier (or filter) but the gain at dc goes to ∞so applications probably limited.

First-Order Highpass Filter

OUT

IN

V 1+sRC=V sRC

But is this a useful circuit?

( )1 OUTV sC+G = V sC

( )1 INV sC+G = V G

OUT

IN

V 1= +V sRC

Requires matched resistors and matched capacitors

Actually uses a concept called “pole-zero cancellation”

Generally less practical than the cascade with an inverter

Two-capacitor noninverting integrator


( ) ( )( )

F

1

Z sT s

Z s= −

s-domain representation

If Z1=1/sC, ZF=R, obtain ( )T s sRC= −




Consider the differential equation ( )dx ty K

dt=

( ) ( )Y s = K s X sTaking the Laplace Transform, obtain ( ) ( )( )

Y sT s = = Ks

X s

Thus, this circuit is an inverting differentiator with a unity gain frequency of K-1 = (RC)-1

K-1 is the frequency where |T(jω)|=1

( )T s sRC= −

Inverting Differentiator

( )T s sRC= −

Differentiator gain ideally goes to ∞ at high frequencies

Differentiator not widely used

Differentiator relentlessly amplifies noise

Stability problems with implementation (not discussed here)

Placing a resistor in series with C will result in a lossy differentiator that has some applications

First-order High-pass Filter

( ) 2

1

RT s = - 1R +sC

2

1

sR C = -1+R Cs

( )( )2

2

1

ωR C T jω = 1+ ωR C

( )T jω

1

1R C

2

1

RR

Applications of integrators to solving differential equations

IN OUT

Standard Integral form of a differential equation

1 2 3 0... ...OUT OUT OUT OUT IN IN INX b X b X b X a X X X= + + + + + + +∫ ∫∫ ∫∫∫ ∫ ∫∫Standard differential form of a differential equation

' '' ''' ' ''1 2 3 1 2 3... ...OUT OUT OUT OUT IN IN INX X X X X X Xα α α β β β= + + + + + + +

Initial conditions not shown

Can express any system in either differential or integral form


LinearSystemXIN XOUT

Consider the standard integral form

1 2 3 0... ...OUT OUT OUT OUT IN IN INX b X b X b X a X X X= + + + + + + +∫ ∫∫ ∫∫∫ ∫ ∫∫

∫ ∫ ∫ ∫INX

∫ ∫ ∫ ∫OUTX

a0 a1 a2a3 am

b1b2

b3 bn

This circuit is comprised of summers and integratorsCan solve an arbitrary linear differential equationThis concept was used in Analog Computers in the past





Take the Laplace transform of this equation

1 2 3 0 1 2 31 1 1... ...2 3 n 2 3 m

1 1 1 1 1s s s s s s s sOUT OUT OUT OUT n IN IN IN IN mb b b b a a a a a= + + + + + + + + + +X X X X X X X X

Multiply by sn and assume m=n (some of the coefficients can be 0)

1 2 3 0 1 2 3... ...n n-1 n-2 n-3 n n-1 n-2 n-3s s s s s s s sOUT OUT OUT OUT n IN IN IN IN nb b b b a a a a a= + + + + + + + + + +X X X X X X X X

( ) ( )1 2 3 0 1 2 3... ...n n-1 n-2 n-3 n n-1 n-2 n-3s s s s s s s sOUT n IN nb b b b a a a a a− − − − − = + + + + +X X

( ) 0 1 2 3

1 2 3

......

n n-1 n-2 n-3

n n-1 n-2 n-3

s s s ss s s s

OUT n

IN n

a a a a aT sb b b b+ + + + +

= =− − − − −

X X





( ) 0 1 2 3

1 2 3

......

n n-1 n-2 n-3

n n-1 n-2 n-3

s s s ss s s s

OUT n

IN n

a a a a aT sb b b b+ + + + +

= =− − − − −

X X

( ) 1 1 0

1 1

......

n n-1

n n-10

s s ss s s +

n n

n

T s α α α αβ β β

−

−

+ + +=

+ + +

This can be written in more standard form

Applications of integrators to filter design

LinearSystemXIN XOUT ( ) 1 1 0

1 1

......

n n-1

n n-10

s s ss s s +

n m

n

T s α α α αβ β β

−

−

+ + +=

+ + +

∫ ∫ ∫ ∫INX

∫ ∫ ∫ ∫OUTX

-β0

-βn-1-βn-2

-βn-3

nαn-1α

n-2αn-3α

0α

Can design (synthesize) any T(s) with just integrators and summers !

Integrators are not used “open loop” so loss is not added

Although this approach to filter design works, often more practical methods are used

End of Lecture 12


01Is

− 02Is+

( )20

2 2 20 0

IT s

s + I s+Iα

−=( ) 0

1 2 20 0

I sT s

s + I s+Iα

−=

( )01OUT1 IN OUT2 OUT1

IX = - X +X +αXs

⎛ ⎞⎜ ⎟⎝ ⎠

02OUT2 OUT1

IX = Xs

⎛ ⎞⎜ ⎟⎝ ⎠

( )OUT1 011 2IN 01 0102

I sX =T sX s + I s+I Iα

−=

( )OUT2 01022 2IN 01 0102

I IX =T sX s + I s+I Iα

−=

This is a two-integrator-loop filter

These are 2-nd order filters

If I01=I02=I0, these transfer functions reduce to


01Is

− 02Is+

( )20

2 2 20 0

IT s

s + I s+Iα

−=

( ) 01 2 2

0 0

I sT s

s + I s+Iα

−=

Consider T1(jω)

( ) ( )20

1 200

jωIT jω

I -ω +jω Iα

−=

( )( ) ( )22

01 22

00

ωIT jω

I -ω + ω Iα=

( )1T jω

This is the standard 2nd order bandpass transfer function

Now lets determine the BW and ωP


( ) 01 2 2

0 0

I sT s

s + I s+Iα

−=

( )( ) ( )22

01 22

00

ωIT jω

I -ω + ω Iα=

Determine the BW and ωP ( )1T jω

( )1 PT jω

( )2

1 PT jω

( )01T jω

ωd

d=

( ) 201

2T jω

ω

d

d=

( )( ) ( )

2

22

2 20

1 2200

ω IT jω

I -ω + ω Iα=

( ) ( ) ( ) ( ) ( )( )( ) ( )

2 22 22

2 222

222 2 2 2 2

0 00 0 0 0122

00

I -ω + ω I I -ω I I -ω IT jω

I -ω + ω I

d

d

α α

ωα

⎛ ⎞− +⎜ ⎟

⎝ ⎠=⎡ ⎤⎢ ⎥⎣ ⎦

0=

To determine ωP, must set

This will occur also when and the latter is easier to work with


( ) 01 2 2

0 0

I sT s

s + I s+Iα

−=

Determine the BW and ωP

( ) ( ) ( ) ( ) ( )( )( ) ( )

2 22 22

2 222

20

22 2 2 2 20 00 0 0 01

2200

I -ω + ω I I -ω I I -ω IT jω

I -ω + ω I

d

d

α α

ωα

⎛ ⎞− +⎜ ⎟

⎝ ⎠= =⎡ ⎤⎢ ⎥⎣ ⎦

( ) ( ) ( ) ( )( )2 22 2222 2 2 2 2

0 00 0 0 0I -ω + ω I I =ω I I -ω Iα α⎛ ⎞

− +⎜ ⎟⎝ ⎠

P 0ω = I

( )( ) ( )2

10 01 P 2 2

00 0

I IT jω

I -I + I αα= =

It suffices to set the numerator to 0

Solving, we obtain

( )( ) ( )22

01 22

00

ωIT jω

I -ω + ω Iα=

Substituting back into the magnitude expression, we obtain

Although the analysis is somewhat tedious, the results are clean

( )1T jω

( )1 PT jω

( )2

1 PT jω

The 2nd order Bandpass Filter


( ) 01 2 2

0 0

I sT s

s + I s+Iα

−= ( )

( ) ( )22

01 22

00

ωIT jω

I -ω + ω Iα=

Determine the BW and ωP ( )T jω

1α

12α

( ) ( ) ( ) ( )( )( ) ( )

2 22 2

2 222

212

22 2 2 2 20 00 0 0 0

2200

I -ω + ω I I -ω I I -ω I

I -ω + ω I

α α

αα

⎛ ⎞− +⎜ ⎟

⎝ ⎠=⎡ ⎤⎢ ⎥⎣ ⎦

H L 0BW = ω - ω = Iα

H L 0ω ω = I

To obtain ωL and ωH, must solve ( ) 121T jωα

=

This becomes

The expressions for ωL and ωH can be easily obtained but are somewhat messy, butfrom these expressions, we obtain the simple expressions



( ) 01 2 2

0 0

I sT s

s + I s+Iα

−=


P 0ω = I ( ) 11 PT jω

α=

( )( ) ( )22

01 22

00

ωIT jω

I -ω + ω Iα=

( )1T jω

( )1 PT jω

( )2

1 PT jω


( )1T jω

12α

1α


( ) 01 2 2

0 0

I sT s

s + I s+Iα

−=


0BW = αIH L 0ω ω = I

( )1T jω

12α

1α

( ) 01 2 2

0

I sT s

s +BWs+I

−=


Often express the standard 2nd order bandpass transfer function as


These results can be generalized

BW = a

Pω b =

( )BPT jω

K

K2

( )BP 2HsT s

s +as+b=


HK=

a


( ) 01 2 2

0 0

I sT s

s + I s+Iα

−=


( )1T jω

12α

1α

( ) 01 2 2

0

I sT s

s +BWs+I

−=


01Is

− 02Is+

Can readily be implemented with a summing inverting integrator and a noninverting integrator


( ) 01 2 2

0 0

I sT s

s + I s+Iα

−=


( )1T jω

12α

1α

( ) 01 2 2

0

I sT s

s +BWs+I

−=


01Is

− 02Is+

• Widely used 2nd order Bandpass Filter• BW can be adjusted with RQ• Peak gain changes with RQ• Note no loss is added to the integrators

P 0ω = I0BW = αI

01I =

RCBW=

RCα

∴


( ) 0BP 2 2

0 0

I sT s

s + I s+Iα

−=

Design Strategy

( )1T jω

12α

1α

The 2nd order Bandpass Filter0Is

− 0Is+

1. Pick C (use some practical or convenient value)

2. Solve expression to obtain R

3. Solve expression to obtain α and thus RQ

01I =

RCP 0ω = I

P1ω =

RC

BW=RCα

BW=RCα

∴

Assume BW and ωP are specified


Exact expressions for BW and ωP are very complicated but ωP≈I0

( )20

2 2 20 0

IT s

s + I s+Iα

−=

The 2nd order Lowpass Filter

01Is

− 02Is+

• Widely used 2nd order Lowpass Filter• BW can be adjusted with RQ but expression not so simple• Peak gain changes with RQ• Note no loss is added to the integrators

( )2T jω

ωP

ω

( )2 PT jω

( )2

2 PT jω BW

01I =

RC

Design procedure to realize a given 2nd order lowpass function is straightforward

Another 2nd-order Bandpass Filter

( )1 1 2 2 3 OUT 2 IN 3V sC +sC +G +G = V sC +V G

11 1 OUTV sC +V G = 0

( )

( )

3 2

2

1 1 1 2 2 3 1 1 2

sR CT s = -

1 1 1s +s + +R C R C R //R R C C⎛ ⎞⎜ ⎟⎝ ⎠

( )

( )

3

22

1 2 3 1

sR CT s = -

2 1s +s +R C R //R R C⎛ ⎞⎜ ⎟⎝ ⎠

1

2BW = R C

( )P1 2 3

1ω =R R //R C

( )BPT jω

K

K2

1

3

RK= 2R

If the capacitors are matched and equal to C

Since this is of the general form of a 2nd order BP transfer function, obtain


( )

( )

3

22

1 2 3 1

sR CT s = -

2 1s +s +R C R //R R C⎛ ⎞⎜ ⎟⎝ ⎠

1

2BW = R C ( )P

1 2 3

1ω =R R //R C

( )BPT jω

K

K2

1

3

RK= 2R

Design Strategy

1. Pick C to some practical or convenient value

2. Solve expression to obtain R1

3. Solve expression to obtain α and thus R3

4. Solve expression to obtain R2

1

2BW = R C

Assume BW, ωP , and K are specified

1

3

RK= 2R

( )P1 2 3

1ω =R R //R C


( )P1 2 3

1ω =R R //R C

( )1 2 3 OUT IN 3 OUTsCV sC +sC +G +G = V sC +V G VH

+

( ) 1OUT

1 1 OUTV sC+G =V sC+V G

H

( )

( )( ) ( )1

3

22

1 2 3 2 3 1

s HR C H-1T s = -

2 1s +s +R C R //R H-1 R //R R C

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞−⎜ ⎟

⎝ ⎠

( )( )1

1 2 3

2BW = R C R //R H-1

−

( )( )1

3

1 2 3

1 HR H-1K =

2R R //R H-1

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞−⎜ ⎟

⎝ ⎠

For the appropriate selection of component values, this is one of the best 2nd order bandpass filters that has been published

( )BPT jω

K

K2

Termed the “STAR” biquad by inventors at Bell Labs

STAR 2nd-order Bandpass Filter

( )

( )( ) ( )1

3

22

1 2 3 2 3 1

s HR C H-1T s = -

2 1s +s +R C R //R H-1 R //R R C

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞−⎜ ⎟

⎝ ⎠

Implementation:

But the filter doesn’t work !

???

STAR 2nd-order Bandpass Filter

( )

( )( ) ( )1

3

22

1 2 3 2 3 1

s HR C H-1T s = -

2 1s +s +R C R //R H-1 R //R R C

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞−⎜ ⎟

⎝ ⎠

Works fine !

Reduces to previous bandpass filter at H gets large

Note that the “H” amplifier has feedback to positive terminal

Implementation:

???

If op amp ideal, OUT

IN

V = HV

Will discuss why this happens later!

ee 230 lecture 12 spring 2010 - iowa state universityclass.ece.iastate.edu/ee230/lectures/ee 230...

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